For the third one by hand, I was just trying to keep track of 1-2 of the first digits for each section.
I broke into sections as follows (^ to mean power here): 9! = (I just did this by hand, 1728x210=362880) , so we take 36. 10! to 19! = approx 1^10 = 1 20! to 29! = approx 2^10 = 1024 = 10 30! to 39! = approx 3^10 = 3, 6, 9, 27, 81, 243, 72.., 216.., 63..., 189.. = 1 40! to 49! = 4^10 = 4, 16, 64, 25.., 10.., 4.. = this looked like a pattern, the tenth one should be 1.. again = 1 50! to 59! = 5^10 = 5, 25, 125, 60.. , 30.., 150.., 75.., 37.., 18.., 90 = 9 60! to 67! = 6^8 = 6, 36, 216, 126.. , 72..., 43.., 25.., 15.. = 1 so actually to correct what I had above, I was really looking to do 9*36 , (not 9*3=27), which gives about 324, or 3 for the first digit. This is just an approximation but it seems to work. On Aug 30, 4:06 pm, kartik sachan <kartik.sac...@gmail.com> wrote: > for 1st question ans will be a,b,e as (125)^1/3 is 5 sum is 8 > for 2nd question we see pattern in 7 power i.e 7,9,3,1 and this pattern > reapeats for inerval of 4 > and second term of of any power of 7>=2 is 4 > so diff is 5 > ans will be 5 > > how to find the ans of third question?? -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
